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Slowly Oscillating Solutions of a Parabolic Inverse Problem: Boundary Value Problems
Boundary Value Problems volume 2010, Article number: 471491 (2010)
Abstract
The existence and uniqueness of a slowly oscillating solution to parabolic inverse problems for a type of boundary value problem are established. Stability of the solution is discussed.
1. Introduction
It is well known that the space of almost periodic functions and some of its generalizations have many applications (e.g., [1–13] and references therein). However, little has been done for
to inverse problems except for our work in [14–16]. Sarason in [17] studied the space
of slowly oscillating functions. This is a
-subalgebra of
, the space of bounded, continuous, complex-valued functions
on
with the supremum norm
. Compared with
,
is a quite large space (see [17–20]). What we are interested in
is based on the belief that
certainly has a variety of applications in many mathematical areas too. In [15], we studied slowly oscillating solutions of a parabolic inverse problem for Cauchy problems. In this paper, we devote such solutions for a type of boundary value problem.
Set . Let
(resp.,
, where
) denote the
-algebra of bounded continuous complex-valued functions on
(resp.,
) with the supremum norm. For
(resp.,
) and
, the translate of
by
is the function
(resp.,
,
).
Definition 1.1.
-
(1)
A function
is called slowly oscillating if for every
,
, the space of the functions vanishing at infinity. Denote by
the set of all such functions.
-
(2)
A function
is said to be slowly oscillating in
and uniform on compact subsets of
if
for each
and is uniformly continuous on
for any compact subset
. Denote by
the set of all such functions. For convenience, such functions are also called uniformly slowly oscillating functions.
-
(3)
Let
be a Banach space, and let
be the space of bounded continuous functions from
to
. If we replace
in (1) by
, then we get the definition of
.
As in [17], we always assume that is uniformly continuous.
The following two propositions come from [15, Section 1].
Proposition 1.2.
Let be such that
is uniformly continuous on
. Then
.
For , suppose that
for all
. Define
by

The following proposition shows that the composite is also slowly oscillating.
Proposition 1.3.
Let . If
and
for all
, then
.
In the sequel, we will use the notations: ,
.
means that
is slowly oscillating in
and uniformly for
;
means that
is slowly oscillating in
and uniformly on
.
Let

be the fundamental solution of the heat equation [21].
2. A Type of Boundary Value Problem
We will keep the notation in Section 1 and at the same time introduce the following new notation:

In this section, we always assume the following: ,
,
,
,
,
,
,
, and
,
.
Let

be Green's function for the boundary value problems [22, 23].
The following estimates are easily obtained:

where (
) are positive and increasing for
and
as
.
To show the main results of this section, the following lemmas are needed. The first lemma is Lemma 3.1 on page 15 in [24].
Lemma 2.1.
Let ,
, and
be real, continuous functions on
with
. If

then

Lemma 2.2.
Let be a continuous function on
. If
,
, and
are nondecreasing and nonnegative on
and

then

where

Proof.
Replacing in the two integrals of (2.6) by the expression on the right hand side in (2.6), changing the integral order of the resulting inequality and making use of the monotonicity of
,
and
, one gets

Apply Lemma 2.1 to get the conclusion.
Lemma 2.3.
Let ,
, and
. Then the problem

has a unique solution , and
is in
and satisfies

where .
One sees that depends on
only and is bounded near zero.
Proof.
The existence and uniqueness of the solution comes from Theorem 5.3 on page 320 in [25].
As in [22, 23], the solution can be written as

So,

By Lemma 2.1, one gets the desired inequality.
Now we show that . As in the proofs of Lemmas 2.1 and 2.3 in [15], one gets
. For
,
with
,

Note that

where is a constant and

So,

By Lemma 2.1, one has

where is a constant. Since
and
are slowly oscillating, the right-hand sides of the inequality above approaches zero as
. This means that
. The proof is complete.
Consider the following problem.
Problem 1.
Find functions
and
such that




One sees that


It follows from (2.24) that

Let , and let
. We have the following two additional problems for
and
, respectively.
Problem 2.
Find functions
and
such that




Problem 3.
Find functions
and
such that




Lemma 2.4.
Problems 1, 2, and 3 are equivalent to each other.
Proof.
The existence and uniqueness of the solution of Problem 2 can be easily obtained from that of the solution
of Problem 1. Conversely, let (
) be the solution of Problem 2. We show that Problem 1 has a unique solution (
). The uniqueness comes from the uniqueness of (2.19)–(2.21). For the existence, let

Obviously, and satisfies (2.22). Also
satisfies (2.21) because
. By (2.23) and (2.27), one sees that (2.20) is true. Finally, we show that
satisfies (2.19) and therefore, along with
, constitutes a solution of Problem 1. In fact,

Thus, we have shown the equivalence of Problems 1 and 2. Replacing (2.34) by the function

the equivalence of Problems 2 and 3 can be proved similarly. The proof is complete.
By Lemma 2.4, to solve Problem 1, we only need to solve Problem 3. By (2.30)–(2.32), we have the integral equation about :

Rewrite (2.33) as

where is determined by (2.37).
One can directly test that Problem 3 is equivalent to (2.37)-(2.38).
Note that for a given , Lemma 2.3 shows that (2.30)–(2.32) (or equivalently, (2.37)) have a unique solution
. Thus, (2.38) does define an operator
. Therefore, we only need to show that the integral (2.38) has a unique solution
and
. That is,
has a fixed point in
. Let

Set , where
. If
, then, by Lemma 2.3,
is in
, and so, by (2.38),
is in
with

Equation (2.37) gives the estimate

Choose such that when
, one has
. It follows that

Choose such that when
, one has

and therefore, .
Let ,
. By (2.38),
. Note that the function
is the solution of the problem

So, by Lemma 2.3, one has

Choose such that for
,
. Now, set
. Then
is a contraction from
into itself, and therefore, has a unique fixed point. Thus, we have shown.
Theorem 2.5.
Let functions ,
,
, and
be as above. Then, for small
, Problem 3 has a unique solution (
) in
with
and
.
Let be the solutions of Problem 3 in
for the functions
,
,
, and
. Set
,
,
, and
. For the stability of the solution, we have the following.
Theorem 2.6.
For , one has

where depends on
,
,
,
,
,
,
, and
.
Proof.
By (2.33),

So,

Note that the function is the solution of the problem

Using a formula similar to (2.37) and Lemma 2.2 for the function , one gets

Applying Lemma 2.2 and (2.48), one gets the desired conclusion with

where

and is majorant of
. One can specially assume that

The proof is complete.
Corollary 2.7.
Under the conditions in Theorem 2.6, the solution of Problem 3 is unique.
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Acknowledgment
The research is supported by the NSF of China (no. 11071048).
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Yang, F., Zhang, C. Slowly Oscillating Solutions of a Parabolic Inverse Problem: Boundary Value Problems. Bound Value Probl 2010, 471491 (2010). https://doi.org/10.1155/2010/471491
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DOI: https://doi.org/10.1155/2010/471491
Keywords
- Differential Equation
- Continuous Function
- Integral Equation
- Partial Differential Equation
- Unique Solution